Calculate the concentrations of Cd ⁺ ², [Cd (CN)₄-²] and CN at equilibrium when they dissolve mole of 0.42 Cd (in the ₃)²⁺ 2.50 M. Given: Kf = 7, 1 x 10 ^16

Your post makes no sense to me as written. I understand what is being asked but I don't understand how the solution was made.

Let's try this... Adding 0.42 mole of Cd^+2 salt into a 2.50M solution of Cyanide salt. The concentration of Cd(CN)4^2- with a Kf = 7.1 x 10^16 will be 0.42mole/L. (100% conversion) Now, the Kd for the Cd-complex will be (7.1 x 10^16)^-1 = 1.4 x 10^-17.

Cd(CN)4^2- <=> Cd^+2 + 4CN^-
Ci: 0.42M 0 0
∆C: -x +x +4x
Ceq: 0.42-x x 4x
~0.42M
Kd = [Cd^+2][CN^-]^4/[Cd(CN)4^2-]
1.4E-17 = (x)(4x)^4/(0.42)
= 256x^5/0.42
Solving for x = [(0.42)(1.4E-14) / (256)]^1/5 = 1.2E-4
[Cd(CN)4^2-] = 0.42M - x ~ (0.42)M
[Cd^+2] = x = (1.2E-4)M
[CN^-] = 4x = 4(1.2E-4)M = (4.8E-4)M

To solve this question, we need to use the solubility product constant (Kf) to calculate the concentrations at equilibrium. The solubility product constant relates to the concentrations of the ions present in a saturated solution.

Step 1: Write the balanced chemical equation for the dissolution of Cd in water:
Cd (in the solid state) ⇌ Cd²⁺ (in solution) + 2e⁻

Step 2: Write the balanced equation for the formation of the complex ion [Cd(CN)₄]²⁻:
Cd²⁺ + 4CN⁻ ⇌ [Cd(CN)₄]²⁻

Step 3: Write the equations for the dissociation of [Cd(CN)₄]²⁻:
[Cd(CN)₄]²⁻ ⇌ Cd²⁺ + 4CN⁻

Step 4: Set up an ICE (Initial, Change, Equilibrium) table for each equation. Let's assume "x" is the change in concentration:

For Cd²⁺:
Initial concentration: 0 M
Change in concentration: +x M
Equilibrium concentration: x M

For [Cd(CN)₄]²⁻:
Initial concentration: 0 M
Change in concentration: +x M
Equilibrium concentration: x M

For CN⁻:
Initial concentration: 0 M
Change in concentration: +4x M (due to the stoichiometry of the reaction)
Equilibrium concentration: 4x M

Step 5: Use the solubility product constant (Kf) to set up the equilibrium expressions:
Kf = [Cd²⁺] [CN⁻]⁴ / [Cd(CN)₄]²⁻

Substituting the equilibrium concentrations from the ICE table, we get:
Kf = x [4x]⁴ / x

Step 6: Simplify the equation using the given values for Kf:
7.1 x 10^16 = 4x⁴

Step 7: Solve this equation for x using algebraic methods:
Divide both sides by 4: 1.775 x 10^16 = x⁴
Take the fourth root of both sides: x = 2.272 x 10⁴ M

Step 8: Substitute the value of x back into the ICE table to find the equilibrium concentrations:
[Cd²⁺] = x = 2.272 x 10⁴ M
[Cd(CN)₄]²⁻ = x = 2.272 x 10⁴ M
[CN⁻] = 4x = 9.088 x 10⁴ M

Therefore, at equilibrium, the concentrations of Cd²⁺, [Cd(CN)₄]²⁻, and CN⁻ are approximately:
[Cd²⁺] = 2.272 x 10⁴ M
[Cd(CN)₄]²⁻ = 2.272 x 10⁴ M
[CN⁻] = 9.088 x 10⁴ M